Question

Q.69 drive an expression for the excess pressure inside a liquid drop

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Answer 1

Explanation:

Answer

consider a s:-

small spherical liquid drop with a radius R.

it has a convex surface so that is pressure P

on the concave side (inside the liquid)

greater than the pressure Po on the convex side

(outside liquid). the surface area of drop is

A=4πR²_________(1)

imagine an increase in radius by an infinity simal

amount dR from the equilibrium value R.

then the differential increase in surface area would be

dA=8πR.dR__________(2)

the increase in surface energy would be equal to the required to increase the surface area

dW=TdA= 8πTRdR________(3)

we assume that dR is also small that the pressure inside remains the same equal to p.

all parts of surface of the drop experience can outward force per unit area is equal to Tu p-po

force during the increase in radius dR

dW= (excess pressure× surface area).dR

= (p-po)×4πR².dR__________(4)

Form Eqs.(3)&(4)

(p-po)×4πR².dR=8πTRdR

$∴ \: p - po = \frac{2T}{R}$

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Answer 2

Answer

→ This gives, P1−P0=r2T which represents the excess pressure inside the liquid drop.

All the best :)